A half-point ko is the common term for a ko in which nothing is at stake beyond the fate of the stone inside the ko.
Despite the name, the miai value of connecting or capturing in a half-point ko is only one third of a point.
This name is only there because it translates the Japanese hanko. I propose minimal ko, since that is what it is.
Half-point ko also translates the Chinese term 半劫 directly.
--unkx80
Bill: I just found a Real Half Point Ko.
Arno: could someone please explain me, why ordinary minimal kos have a value of 1/3? I haven't got the knack of miai counting yet.
Bill: Hi, Arno! :-)
The difference between winning and losing the ko is 1 point. There are 3 moves between the winning positions, 1 move for each player to win the ko and 1 move to take the ko. So each move is worth 1/3 point, on average. Es claro?
Arno: while formulating the next question I guess I got the solution, which I write down for other people like me: Bill counts three moves, because in the following position ...
... if white plays, she has invested one move more than black (i.e. 1 white move, 0 black moves). If black plays, he has to capture and connect (=2 moves) while white does not play there (=0 moves). Therefore, black played two more moves than white. A possible ko-threat play by White is not counted, because it is outside this position.
Thus the difference of the possible outcomes is 3 moves and 1 point value, thus a miai value of 1/3.
Did I get it right? :o)
Bill: Well put, Arno! :-)
KarlKnechtel: Interesting... I came up with the 1/3 value in an entirely different way. First, let's see if I've got the basics right:
In this diagram, white has half a point, under the assumption that it's even probability that each player will be the next to play locally (in turn based on the assumption that there are roughly equal-value plays elsewhere on the board). Therefore, whoever plays at 'a' gains half a point for that move. That's about the limit of my confident understanding of miai counting ^^; But assuming I have the basic idea right, I apply it to the minimal ko as follows:
Now there is an even chance that white will play next. Define the result to be net 0 points difference. The alternative is that black eventually wins the ko, and gets +1 point as a result.
If, at each move, it is an even chance that either white or black will play again locally, the chance that black wins the ko immediately is 1/4; that white wins immediately is 1/2; the remaining 1/4 of the time, black plays and then white plays, so that the original situation is restored. Thus, when we sum the infinite series, we get a total chance of 2/3 that white wins, and 1/3 that black wins.
Thus the current result gives white a disadvantage of 1/3 point from the 'reference' position where white has won the ko. Thus, playing locally is worth 1/3 point for white. Symmetrically, it is worth 1/3 point for black, since the black play would shift the value from 1/3 to 2/3 in black's favour (the value would become 2/3 since it's 1/3 less than the +1 value for black winning the ko - 1/3 less by the same argument as above).
This way I don't have to think about counting moves or amortizing the value of a play over the moves in the sequence (which I find a bit counterintuitive). But I'm not sure it's any easier to understand ;) (Actually, I just make it sound complicated, I think.)
MK In fact this reasoning nearly convinced me. Could you please make the 'infinite series' part clearer?
victim 1/2 - 1/4 + 1/8 - 1/16 + ... = 1/3.
Another way to come to the same result: Let's call the position where Black can capture or White can connect, B. If White connects, we get zero. If Black captures, we get a position we call W, where Black can connect or White can capture. So B = 0 * 1/2 + W * 1/2, if the probability is 1/2 for each outcome.
Similarly, for the next move, W = B * 1/2 + 1 * 1/2. Now we have two equations with two variables.
B = W/2
W = B/2 + 1/2 = W/4 + 1/2
W*3/4 = 1/2
W = 2/3
B = W/2 = 1/3
By the way, this is not as useless as it seems to be at first sight. The computation works the same way for a bigger ko. So, as a first estimate of how big capturing a ko is, compute the difference between "White wins" and "Black wins" and multiply by 2/3. If you don't have much time, play a gote move bigger than that rather than playing the ko.
Of course the number changes if it's a two-stage ko.
Capturing or connecting here is worth the same as two points in gote, as an approximation. Use the same methods as above to calculate that.
RafaelCaetano Karl, I agree that you make it seem more complicated than it is. :-) I don't see why introduce probabilities and infinite series here.
I find Bill and Arno's reasoning convincing enough. But since some people don't, I'll try a simpler approach. Actually I read it somewhere else a long time ago, probably r.g.g (maybe by Bill?). Consider the following position.
Black plays first. White connects, since it's no use to fight the ko. Black 3 also connects (below 1), since again it's useless to try to win both kos. Finally White connects. Black got a stone, that is, 1 point.
White plays first. Again, after Black captures with 2, White simply connects the other ko. Black 4 below 2. Again Black got 1 point.
So the value of the position above is 1. Therefore, the value of each of the 3 kos is 1/3.
Is this reasoning correct? It looks so simple that I feel something is missing.
victim It's fine. Of course one has to remember that it's still an approximation.